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Answers_OldTests - MGMT 36100 Test 1 - A Question 2 Answers...

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Unformatted text preview: MGMT 36100 Test 1 - A Question 2 Answers to Old Tests These are not complete solutions, some steps have been omitted. Part A: Stage 1: 15 [units / min], Stage 2: 18 [units / min], Stage 3: (20 [units / min] Part B: Productivity: [150 units/ worker /hour] Part C: If the input rate is 20, it will exceed the process capacity and the process will not be stable. Part D: a. We should select P4 and M3 because stage 3 above has the highest capacity. Adding a machine at stage 3 will not help improve process capacity. b. Capacity of P4 should be 5 [units / min]. Capacity of M3 should be 2 [units / min]. This will make capacities of all stages equal at minimum cost. Question 3 Part A: Claim: TPT < 6 moths.TPT = 18 months. Claim is not justified Part B: Total time per worker per day (in sec.): 27000 [sec / shift]. See calculations in the table below. Part C: Hourly capacity of Stage 3: 14 [units / hr] What is the minimum TPT? Part B Explain briefly. X will be completed in 20 min, Y Time needed # of workers in 25 and Z in 6. Since all 3 are needed in stage 4, min TP = 43 minutes. D1 187,200 [sec / shift] = 6.93 [workers] ≈ 7 D2 D3 156,000 [sec / shift] 195,000 [sec / shift] = 5.78 [workers] ≈ 6 = 7.22 [workers] ≈ 8 Question 4 Part A: Rank positional weight in the table to the right. Part B: target CT = 53 [sec / unit] (We must round it down to produce at least 470 units). Part C: LPT order 2 1 6 3 5 Task Weight 1 60 2 75 3 85 4 25 5 30 6 40 7 10 47 Station Tasks I II III IV Chase Basic Plan Inventory Adjust. 1 Inventory Adjust. 2 Inventory Adjust. 3 Inventory P1 3000 0 3000 0 3000 0 3000 0 2 1, 3 6, 5 4, 7 P2 2000 0 3200 1200 3700 1700 3500 1500 Part D: Effective CT = 60 sec/unit. Efficiency = 75% 2 45 1, 3 60 6, 5 50 4, 7 25 Question 5 Part A: Part B: Capacity exceeded in P4 and P6. Adj. 1: Move 1200 from P4 to P2. Capacity exceeded P6. Adj. 2: Move 1000 from P6 to P5, move 500 from P6 to P2. Inventory exceeded in P2 and P3. Adj. 3: Reduce 200 from P2. This will produce shortage in P6. The plan is now acceptable. Cumulative Station time 45 35, 60 30, 50 15, 25 P3 3500 0 3500 1200 3500 1700 3500 1500 P4 5200 0 4000 0 4000 500 4000 300 Eligible tasks 2,1,3 / 1,3 1,3 / 3,4 / 6,5,4 6,5,4 / 5,4 / 4 4 / 7 P5 2500 0 2500 0 3500 1500 3500 1300 P6 5000 0 5000 0 3500 0 3500 ‐200 Test 1 - B Question 2 Part A: CT at stage 1 = 3.33 [minutes / unit]. CT at stage 2 = 3.75 [minutes / unit] Part B: process productivity = 3.2 units per labor per hour. Part C: minimum TPT = 13.5 min. Part D: For both choices, the bottleneck is shifted to stage 1. Choose cheaper alternative ‐ replace P3 with P4. Answers_OldTests.doc Page 1 Question 3 Part A: Flow time = 588 1.96 years! Obviously, this is far greater than 9 months. Hence we disagree. Part B: Task 1234 (1) A single line will produce 2016 per day. No. of lines needed: 3. Time (sec). 24 20 14 11 (2) Per unit labor cost = 3024 / 6000 = 0.504 ≈ 50 cents. (3) Bottleneck: station 1 (35 sec.). Line capacity = 1440 items/day. Question 4 Part A: Target cycle time = 18 sec. / unit Part B: Lower bound = 8 stations. Part C: Tasks in LPT order 1 3 12 2 10 4 11 5 7 9 8 6 Part D: Line efficiency = 90.71% Layout not shown. Question 5 Part A: (1) Requirement of EE in Jan. = 1920 Station Tasks Cum station time I 1 16 II 3 16 III 2,6 12, 17 (2) See table to the right. Final plan Inventory 4800 250 5400 450 5400 5400 50 0 Requirement (std units) Item Nov. Dec. Jan. EE 1600 2400 1920 SS 1200 1800 1000 DD 720 960 1280 Total 3520 5160 4200 Test 2 -A Question 2 Part A: See table to the right. Part B: See table below . SS Period 0 1 2 3 4 5 6 7 Demand 0 90 0 80 80 50 90 110 ‐ ‐ 200 ‐ ‐ 200 ‐ FOQ 200 20 Inventory 120 30 30 150 70 20 130 20 Lot size price / unit H ($/unit/year) Q0 Lot size Q* Annual costs Purchase $ Ordering $ Holding $ Total $ X Pr(X) F(X) 450.00 2 * (1000/2) = 1000.00 20894.44 20800.00 50 100 150 200 250 300 350 0.05 0.15 0.2 0.3 0.2 0.08 0.02 0.05 0.20 0.40 0.70 0.90 0.98 1.00 < 1000 10.00 2.0 447.2 450.0 1000 and up 9.80 2.0 SQRT[ 2 * 2000 * 100/2 ] = 447.2 1000 Question 3 Part A: C = $30, R = $50, V = $25. SL = 0.8, Q0 = 1800, P(Q)= $28,000 Part B: SL = 0.75, Q0 = 250, Expected sold = 182.5. E xpected unsold = 67.5, Expected profit = $4800. 20000.00 2000 * 9.8 = 19600 .00 444.44 100 * (2000 / 1000) = 200.00 Question 4 Part A: (i) Q0 ≈ 6000 units. Run time: 40 days. (ii) Lot size = 4500 units. Annual inventory cost: $2110 Part B: ‐ See Table. 40 = 60 + (350 + 485) ‐ 855 Day Opening IL Opening IP Receipt Order Withdrawal Closing IL Closing IP 1 250 600 0 190 60 410 2 60 410 350 135 275 275 3 275 275 0 485 140 135 620 4 135 620 0 150 ‐15 470 5 ‐15 470 485 150 320 320 6 320 320 0 440 140 180 620 7 180 620 0 140 40 480 Total 1045 Page 2 Answers_OldTests.doc Question 5 Part A: Average X bar 10.1875 R 22.375 A2 0.337 D3 0.184 D4 1.816 Chart X‐bar R LCL CL UCL 2.65 10.19 17.73 4.12 22.38 40.63 Points X bar eliminated R 12 14 16 9 14 Part B: 92.7% components will be accepted. Test 2 - B Question 2 Part (A): i. QO = 4981.2 ≈ 5000 ii. Run time =1 day iii. Cycle time = 20 days. Number of production runs per year = 15 iv. AIC(5000) = $1987.5 Vendor Q 0 Lot size Q* Part (B): X 1732 1800 Y 1732 4500 Question 3 Part (A): Buy 180 in P1, 60 in P2 and P3, 60 in P5, 120 in P6, 60 in P8 and P9; 120 in P10. Part (B): (i) Avg. Demand = 83.80 units (ii) SL = 0.75. Buy 95 units. Avg. no. sold = 77.4. Expected profit = 4293 (iii) Avg. no. sold = 78.55 Annual costs Purchase: $ 1080000 1058400 Ordering: $ 1500 600 Holding: $ 1620 4050 Total: $ 1083120 1063050 Question 4 Part (A): Z0.93 = 1.48. ROP ≈ 1555 Part (B): SL = 97.83% Day Opening Opening Receipt Order With‐ Closing Closing Part (B): I.L. I.P. draw I.L. I.P. Closing IP (day 6) = Opening IP (day 2) + Orders 170 570 0 30 430 1 140 from 2 to 6 – Withdrawals from 2 to 6. 660 = 430 + (505 + 485) – (135 + 140 + 190 + 150 + 30 430 400 295 295 2 135 140)= 430 + 985 – 755 = 660 295 295 0 505 140 155 660 3 155 660 0 ‐35 470 4 190 Question 5 ‐35 470 505 320 320 5 150 (i) UCL = R_bar * D4. D4 = 2.282. From tables, sample size n = 4 285 320 0 480 140 180 660 6 (ii) R_bar = 17.3. For R chart, LCL = 0 * 17.3 = 145 660 0 40 520 7 140 0. UCL = 2.115 * 17.3 = 36.59 905 985 1035 (iii) CPK = 1/3. Probability of acceptance = 84% Total Test 3 (Final) - A Question 2 Part A ‐ (i) TPT for Job 1: 3, Job 2: 4, Job 3: 3, Job 4: 4, Job 5: 4. Average TPT = 3.6 minutes Answers_OldTests.doc WIP contributions W1 1 W2 1 D1 5/6 Buffer 1/6 Total 3 Page 3 (ii) Output rate = (3 / 3.6) * 60 = 50 [units/hr] Part B ‐ Each machine can produce 1.5 units of standard product per hour. Each machine is available for 21 * 8 = 168 hours per month and will produce 168 * 1.5 = 252 units So we need 1540 / 252 = 6.111 ≈ 7 machines. Product AA BB CC DD EE Prod. rate Monthly [Units/hr] demand Demand in standard units (CC) Part C ‐ weight of task 2 =88 i. RPW 2 1 5 4 8 3 7 6 10 11 9 12 ii. St. Selected Cum. time Eligible I 2, 1, 3 8, 20, 26 2,1,3 / 1, 5*, 3 / 5, 4, 3/ 5, 4, 6 II 5, 4, 7 6, 16, 25 5, 4, 6/ 4, 8**, 6 / 8, 7, 6/ 8, 6 III 8, 6, 10 8, 15, 27 8, 6/ 6, 10, 11 / 10, 11, 9 / 11, 9 IV 11, 9, 12 10, 19, 28 11, 9 / 9 / 12 *You can remove task 5 from eligible list for station 1 ** You can remove task 8 from eligible list for station 2 2.0 80 (80 / 2.0) * 1.5 = 60 3.0 200 (200 / 3.0) * 1.5 = 100 1.5 480 (480 * 1.5) * 1.5 = 480 4.0 320 (320 * 4.0) * 1.5 = 120 1.0 520 (520 * 1.0) * 1.5 = 780 Aggregate Requirement 1540 Layout not shown. Line efficiency = 94.6%. Question 3 Part A – D = 12000 units/year, S = $200 per order, H = $3.25/unit/yr Q0 = 1215.3. We should try Q = 1000 and Q = 1250. AIQ(1000) = $4025. AIQ(1250) = $3951.25. Therefore Q* = 1250 and AIC(Q*) = $3951.25 AIQ(1500) = $4037.5 . It will therefore cost 86.25 dollars more annually …. (4037.5 – 3951.25) Part B – (1) Q0 = SQRT(2 * S * D/H) * SQRT[ p/ (p ‐ u) ] For W: Q0 = 1883.44. Q* = 1900 will give run time of 19 days. CT = 1900 / 36 = 50 days. For Z: Q0 = = 1500. No rounding is necessary. Q* = 1500 will give run time of 30 days. T = 1500 / 30 = 50 days. (2) Since the cycle time is same. We can use the following. Run Z for 30 days followed by followed by setup time (S = 1/2 day). Then run W for 19 days followed by followed by setup time (S = 1/2 day). Or you can start with W then produce Z. X S Y S Part C ‐ For the Q system: SS = 466. For the P system: SS = 699. Increase in SS = 233 units Part D ‐ (1) Cs = 50 ‐ 30 = 20. Ce = 30 – 25 = 5. SL = 0.8. Q0 = 1600 (2) P(Q) = $27750 Part E ‐ 1. The process is capable and current Cpk is below acceptable level. We can make adjustments to the process. 2. 0.75 = Cp = (UTL – LTL)/(6σ). Therefore (UTL – LTL) = 4.5σ Or (UTL ‐ μ ) = (μ ‐ LTL) = 2.25σ L μ U Prob. of accepting = 0.9756. Answers_OldTests.doc Page 4 Question 4 Part A ‐ X Y A B L = 2 FOQ 200 SS = 40 C L = 2 POQ 2 SS = 50 POR POR POR GR SR OH PR POR GR SR OH PR POR 0 0 0 0 ‐ ‐ 1050 ‐ ‐ ‐ ‐ 1070 ‐ ‐ 1 30 20 500 1020 200 230 0 400 1430 1200 840 0 1580 2 3 40 50 100 50 0 180 100 410 ‐ ‐ 130 120 0 400 200 400 720 1230 ‐ ‐ 120 470 0 1580 0 1580 4 40 150 0 150 ‐ 170 200 0 420 ‐ 50 0 0 Part B: Diagram not shown 5 6 7 90 50 35 20 60 0 200 0 130 420 60 260 ‐ ‐ ‐ 150 90 230 400 0 400 400 0 0 1310 270 235 ‐ ‐ ‐ 320 50 50 1580 0 235 235 0 0 Item Child Y X S M G F E B ‐ S M, Y ‐ Y M, G F, Y F 0 0 0 0 0 0 0 0 Level 1 2 1 1 2 1 2 1 Final 3 3 0 1 2 2 1 0 0 Part C ‐ Total cost for excess inventory = $9.75 (a) Net revenue without notional penalties: $136 (b) Net revenue with notional penalties: $120.25 . Test 3 (Final) - B Question 2 Part A: 1. See table. 2. 6 workers are needed.Productivity = 50 units per labor hr. 3. We should add a machine at stage 3 such that the capacity is raised to the next higher value (330). Cycle time = 120 sec / unit. Stage 1 Stage 2 Stage 3 Stage 4 Bottleneck Capacity = 360 units/hr Capacity = 450 units/hr Capacity = 300 units/hr Capacity = 330 units/hr Stage 3 Part B: To produce 3 units of X we need 6 minutes. To produce 2 units of Y we need 10 minutes. To produce 1 unit of z we need 8 minutes. Total time used above is 6 + 10 + 8 = 24 min. X: Units produced= 60. Y: Units produced= 40. Z: Units produced = 20. Part C: Time Task 1 2 3 127 Weight 89 86 67 Part D: Chase Capacity Demand Basic Plan End Inventory Adjust. 1 End Inventory Adjust. 2 End Inventory 0 1 5000 4250 4250 0 4250 0 4250 0 2 5000 4000 4000 0 4300 300 4200 200 3 5000 4700 4700 0 5000 600 5000 500 4 5000 5600 5600 0 5000 0 5000 ‐100 5 5000 4600 4600 0 4600 0 4700 0 Basic plan: Capacity exceeded by 600 in P4 Adjustment 1: From P4, move 300 units to P3 and 300 to P2. Now inventory in P3 exceeds 500. Adjustment 2: Move 100 units from P2 to P5. Plan is now acceptable. 0 0 Answers_OldTests.doc Page 5 Question 3 Part A: Part B: R = 200, C = 120, V = 120 SL = 1. Buy 250 units. P(250) = $14000 Part C: Z0.94 = 1.56. Q system: ROP = 1156 ≈ 1160. Hence SS = 160 P system: OUL = 2484 ≈ 2490. Hence SS = 240. Safety stock will go up by 240 – 160 = 80 units. Part D: 10 = 190 + 450 – (80+110+130+90+110+50+60) = 190 + 450 ‐ 630 Part E: Cp = 0.833 Pr (accepting) = 0.933 Question 4 Part A: Range of Q C H Q0 Q* Annual costs purchase price ordering cost holding cost Total annual cost Up to 800 $30/unit 10 547.72 500 Up to 800 $30/unit 10 547.72 600 800 and over $29.50/unit 10 547.72 800 $177,000.00 $1,875.00 $4,000.00 $182,875.00 $180,000.00 $180,000.00 $3,000.00 $2,500.00 $2,500.00 $3,000.00 $185,500.00 $185,500.00 Day Opening Opening Receipt Order With‐ Closing Closing I.L. I.P. draw I.L. I.P. 1 190 640 0 0 80 110 560 2 110 560 450 0 110 450 450 3 450 450 0 0 130 320 320 4 320 320 0 450 90 230 680 5 230 680 0 0 110 120 570 6 120 570 0 0 50 70 520 7 70 520 0 0 60 10 460 8 10 460 450 0 70 390 390 Total 700 Part B: M N X Y L=3 FOQ300 SS=50 Z L=2 POQ 2 SS=200 POR POR POR GR SR OH PR POR GR SR OH PR POR 1 100 80 0 460 600 200 340 0 300 780 0 600 2890 2270* 2195 0 2 30 60 200 810 600 130 0 900 2690 800 200 0 0 3 80 40 200 920 900 110 0 300 1520 0 875 2195 1020 4 5 75 40 0 60 0 200 225 840 0 0 185 245 300 900 300 0 675 900 0 0 200 320 0 1020 0 135 6 90 30 0 330 0 215 300 0 120 0 200 0 0 7 55 80 0 325 0 190 300 0 135 0 200 135 0 8 0 0 0 0 0 190 0 0 0 0 200 0 0 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 Final 2 1 1 3 3 3 3 3 3 1 1 0 0 2 1 Distributor Item Children Level A L, H1, H2 0 B ‐ 0 CA A, SW 0 H1 ‐ 0 H2 ‐ 0 L ‐ 0 Q ‐ 0 R ‐ 0 SC B, CA, R, V 0 ST A, Q 0 SW ‐ 0 V ‐ 0 Part C: After step 1 Brewery Wholesaler Retailer After delivery Party Brewer Distributor Wholesaler Retailer St. 1 8 St. 2 10 Ware St. 1 10 St. 2 6 St. 3 12 Ware St. 1 10 St. 2 6 St. 3 19 Ware St. 1 6 St. 2 9 St. 3 10 Ware 12 St. 1 8 St. 2 0 Ware 10 St. 1 10 St. 2 6 St. 3 3 Ware 9 St. 1 10 St. 2 6 St. 3 4 Ware 15 St. 1 6 St. 2 9 St. 3 0 Ware Quantities Inventory 0 3 4 0 Backlog 6 0 0 0 Costs Inventory cost 0 4.5 6.0 0 Backlog cost 15 0 0 0 Answers_OldTests.doc Page 6 ...
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This note was uploaded on 12/14/2010 for the course MGMT 361 taught by Professor Panwalker during the Spring '10 term at Purdue.

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